#### Need solution for RD Sharma maths class 12 chapter 21 Differential Equation exercise Fill in the blank question 19

$y=\frac{1}{4}x^{2}+Cx^{-2}$

Hint:

Substituting the values

Given:

$x\frac{\mathrm{d} y}{\mathrm{d} x}+2y=x^{2}$

Solution:

On dividing the given differential equation by x, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}+2\frac{y}{x}=x$

So, the differential equation is a linear differential equation, and we can observe that the given differential equation is in the form

$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$

where,

$P=\frac{2}{x} \, and\, Q=x$

Here we get the integrating factor as

$I.F=e^{\int Pdx}$

By substituting the values,

$=e^{\int \frac{2}{x}dx}$

We get,

\begin{aligned} &=e^{2log\, x} \\ &=e^{log\, x^{2}} \\ &=x^{2} \end{aligned}

Here, the general solution can be written as

\begin{aligned} &Y(I.F)=\int Q(I.F)dx+C \end{aligned}

Substituting the values we get

\begin{aligned} &y(x^{2})=\int x(x)^{2}dx+C \\ &x^{2}y=\frac{x^{4}}{4}+C \end{aligned}

Dividing the entire equation by x2

\begin{aligned} &y=\frac{x^{2}}{4}+\frac{C}{x^{2}} \\ \Rightarrow &y=\frac{1}{4}x^{2}+Cx^{-2} \end{aligned}

\begin{aligned} &y=\frac{1}{4}x^{2}+Cx^{-2} \end{aligned}